Relationship Between The Z-Transform And The Laplace Transform Below I've quoted Wikipedia's entry that relates the Z-Transform to the Laplace Transform.  The part I don't understand is $z \ \stackrel{\mathrm{def}}{=}\ e^{s T}$; I thought $z$ was actually an element of $\mathbb{C}$ and thus would be $z \ \stackrel{\mathrm{def}}{=}\ Ae^{s T}$ (but then it would be different to the Laplace Transform...).  I don't understand why the Z-Transform is not defined as:
$$
X(z) = \mathcal{Z}\{x[n]\} = \sum_{n=-\infty}^{\infty} x[n] e^{-\omega n}
$$
or something like that.

Z-transform
The unilateral or one-sided Z-transform is simply the Laplace transform of an ideally sampled signal with the substitution of $$
    z \ \stackrel{\mathrm{def}}{=}\ e^{s T} \  $$
    where $T = 1/f_s \ $is the sampling period (in units of time e.g., seconds) and $f_s \ $is the sampling rate (in samples per second or hertz)
Let $$
    \Delta_T(t) \ \stackrel{\mathrm{def}}{=}\ \sum_{n=0}^{\infty} \delta(t - n T)  $$ be a sampling impulse train (also called a Dirac comb) and $$
    \begin{align} x_q(t) & \stackrel{\mathrm{def}}{=}\ x(t) \Delta_T(t) = x(t) \sum_{n=0}^{\infty} \delta(t - n T) \\ & = \sum_{n=0}^{\infty} x(n T) \delta(t - n T) = \sum_{n=0}^{\infty} x[n] \delta(t - n T) \end{align}  $$ be the continuous-time representation of the sampled x(t) \
$$
    x[n] \ \stackrel{\mathrm{def}}{=}\ x(nT) \ $$ are the discrete samples of x(t) The 
Laplace transform of the sampled signal x_q(t) \ is $$
    \begin{align} X_q(s) & = \int_{0^-}^\infty x_q(t) e^{-s t} \,dt \\ & = \int_{0^-}^\infty \sum_{n=0}^\infty x[n] \delta(t - n T) e^{-s t} \, dt \\ & = \sum_{n=0}^\infty x[n] \int_{0^-}^\infty \delta(t - n T) e^{-s t} \, dt \\ & = \sum_{n=0}^\infty x[n] e^{-n s T}. \end{align}  $$ This is precisely the definition of the unilateral Z-transform of the discrete function $x[n] \ $. $$
    X(z) = \sum_{n=0}^{\infty} x[n] z^{-n}  $$ with the substitution of $z \leftarrow e^{s T} \ $.
Comparing the last two equations, we find the relationship between the unilateral Z-transform and the Laplace transform of the sampled signal: 
$$
    X_q(s) = X(z) \Big|_{z=e^{sT}}
$$ The similarity between the Z and Laplace transforms is expanded upon in the theory of time scale calculus.

(Source: http://en.wikipedia.org/wiki/Laplace_transform#Laplace.E2.80.93Stieltjes_transform)
Here: http://en.wikipedia.org/wiki/Z-transform it says that $z \in \mathbb{C}$.
 A: There is a good reason to use $z$ instead of $e^{sT}$. Before starting with any analysis, let me remind you that in analysis of signals and systems we are interested in analyzing the frequency spectrum of the signal, i.e. the Laplace transform on the imaginary line $s = jw$. And since your signal $x[n]$ is discrete, then its frequency spectrum its periodic, so its more general define $s=jwT$.
Now, let $X(z) = \mathcal{Z}\{x[n]\}$ of a causal or non-causal discrete signal $x[n]$, i.e. 
$$ X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}. $$
Since $z\in\mathbb{C}$ we have $z = |z| e^{j\arg z}$. Without loss of generality we rewrite $|z| = r$ and $\arg z = wT$, i.e. $z=r e^{jwT}$ (note that not necessarily $r=1$). Then
$$ \begin{aligned}
X(z) &= \sum_{n=-\infty}^{\infty} x[n] z^{-n}\\
     &= \sum_{n=-\infty}^{\infty} x[n] (r e^{jwT})^{-n}\\
%     &= \sum_{n=-\infty}^{\infty} (x[n] r^{-n}) e^{-njwT}\\
     &= \sum_{n=-\infty}^{\infty} (x[n] r^{-n}) (e^{jwT})^{-n},
\end{aligned} $$
which implies $X(z) = \left. \mathcal{L}\{x[n]r^{-n}\} \right\rvert_{s=jwT} = \mathcal{F}\{x[n]r^{-n}\}$. As a consequence, $X(z)$ is a Fourier transform more generic than the Fourier transform $X(e^{jwT}) = \mathcal{F}\{x[n]\}$ of our signal of interest.
So, if the convergence radius of $X(z)$ is less than unity then $X(e^{jwT})$ does not exist and therefore its Fourier transform does not either, which represents a problem because there are many signals with this problem of convergence, e.g. non-causal signals such as a digital image filter. Therefore, it is convenient (and even necessary in non-causal signals) to use the Z-transform.
Or informally, use $z$ instead of $\left. e^{sT} \right\rvert_{s=jwT} = e^{jwT}$ whenever you can.
We also recommend to see this link about radius convergence.

At this point, it is clear that the Z-transform has the same objective as the Laplace transform: ensure the convergence of the transform in some region of $\mathbb{C}$, where the Z-transform does it for discrete signals and Laplace transform for continuous signals.
A: Why $z = e^{sT}$ chosen?
If we define $z = e^{sT}$, then the Z-transform becomes proportional to the Laplace transform of a sampled continuous-time signal.

Taken from this slide
https://ccrma.stanford.edu/~jos/Laplace/Laplace_4up.pdf


Why would we want it to be proportional in the first place?
Simple, if we know the correct relation between both of them, then we could working interchangeably between both of the domains. The mapping will be correct.
For example, we may already have a model of perfectly designed continuous domain controller to control our continuous plant. By converting it into the equivalent discrete controller and also including the effects of sampling and zero order hold we could implement it on a digital computer. For flexibility reason and cost effective lets say.
Unfortunately the relation is nonlinear, so it's quite difficult to work with it interchangeably. Commonly, we approximate it by using the first order Taylor Expansion of it, such as Bilinear transform (also known as Tustin's method).

That is the most intuitive reasoning as far as I know.The definition provide a bridge to connect both of the transform theories.
